Q1 of 80 Page 56

Solve the following equation.

Given equation:

(Transposing to LHS and to


RHS)


(LCM of 8 and 16 is 16; of 7 and 14


is 14)




(Taking out 7 common in numerator of RHS)



LCM of 2 and 16 is 16.


(Multiplying by 16 on both sides)


t – 5 = 8 (t – 1)


t – 5 = 8t – 8 (Removing bracket)


t – 8t = -8 + 5 (Transposing 8t to LHS and 5 to RHS)


-7t = -3


t = (Transposing -7 to RHS)


t =


The solution of the given equation is t = .


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